How long was the ball in flight?

[Miike012]

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-fall acceleration on the moon is of its value on earth. Suppose he hit the ball with a speed of 29 at an angle 26 above the horizontal.


How long was the ball in flight?

First I found out that the vertical component of the initial velocity is 29*sin(26 deg) = 12.7.

Next I used the following equation,
V = V₀ + a*t, where a = -9.8/6.

29*sin(26 deg)*6/9.8 = t = 7.8 sec. Is this not the correct answer?

 

[Curious3141]

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-fall acceleration on the moon is of its value on earth. Suppose he hit the ball with a speed of 29 at an angle 26 above the horizontal.


How long was the ball in flight?

First I found out that the vertical component of the initial velocity is 29*sin(26 deg) = 12.7.

Next I used the following equation,
V = V₀ + a*t, where a = -9.8/6.

29*sin(26 deg)*6/9.8 = t = 7.8 sec. Is this not the correct answer?

No, what you've calculated is the time till the ball comes to instantaneous (momentary) rest in the vertical direction (it still remains moving horizontally). What you need to calculate is the time for the ball to reverse its vertical velocity till it's the same magnitude as the initial one but opposite in direction (downward).

Alternatively, since the ball describes a perfect parabolic trajectory, you could just observe that the time you calculated occurs exactly midway in flight (the apex of the parabola), so the total flight time is twice that.

 

[Miike012]

thank you.